BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Game show

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

Game show

by DanaJ » Tue Nov 30, 2010 1:57 pm
Source: Veritas Prep

On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $1, the second $100, and the third $1000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $1000?

(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2

Experts: Only Veritas Prep experts, please!
Top
Source: — Problem Solving |
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Wed Dec 01, 2010 4:02 am
DanaJ wrote:Source: Veritas Prep

On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $1, the second $100, and the third $1000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $1000?

(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2

Experts: Only Veritas Prep experts, please!
IMO D

The contestant must get the key correct to the $1000 box in order to win more than $1000.

P($1000) = 1/3

Out of rest two boxes he must have the correct key to at least one to get more than $1000.

P($100) = 1/2 however as this attempt is success the other attempt P($1) is bound to succeed.
P($1) = 1

Thus P(Winning > $1000) = (1/3)*(1/2)*1 = 1/6

So I am not sure about the other case where we consider P($1) = 1/2 and P($100)=1 as both are equally likely.

Thus, 2*1/6 = 1/3

Not sure :(
If the problem is Easy Respect it, if the problem is tough Attack it
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 307
Joined: Sun Jul 11, 2010 7:52 pm
Thanked: 36 times
Followed by:1 members
GMAT Score:640

by limestone » Wed Dec 01, 2010 4:28 am
[spoiler]IMO: C[/spoiler]

This is somewhat similar to sibling problems.

The desired outcome is the contestant to win more than $1000.

So the key picked for the box that contains $1000 must match. However, the contestant needs to win more than $1000. So another key picked, either for the $100 box or $1 box, must match. And this also leads to the matching of the final box too (As the first two boxes have already had matched keys).

All three boxes must have their respectively matched keys.

My approach:
Pick the first box (whatever it is, never mind 'bout it), possibility for this box to have its matched key : 1/3 (three keys remain)
Pick the second box, possibility for 2nd box to have its matched key: 1/2 (two keys remain)
The final box, possibility for it to have a matched key is absolute, or 100%, or 1.

Total possibility: 1/3*1/2*1 = 1/6

Correct me if I'm wrong.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
Top
User avatar
GMAT Instructor
Posts: 2193
Joined: Mon Feb 22, 2010 6:30 pm
Location: Vermont and Boston, MA
Thanked: 1186 times
Followed by:512 members
GMAT Score:770

by David@VeritasPrep » Wed Dec 01, 2010 4:29 am
Here is the official answer and explanation. You will see that it agrees with the above.

Correct answer: (C)

Solution: This tricky probability question is best solved by calculating the total possibilities and then determining which ones are favorable. There are 3! (or 6) ways that the 3 keys could be assigned. We will call the first key A, the second key B, and the third key C. The six unique assignments are:
ABC, ACB, BAC, BCA, CAB, CBA

Therefore, we know that the total number of possible key distributions is 6. How many are favorable? Interestingly, there is only one favorable outcome that will win more than $1000: when all three boxes are assigned the correct keys. To win more than $1000, at least two of the three boxes must open. If two boxes open, the third must as well. Given that there are six unique key distributions, only one will open all three boxes (i.e., ABC and CBA can't both open all three boxes). Thus the probability of winning more than $1000 is 1/6.
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Wed Dec 01, 2010 4:35 am
DanaJ wrote:Source: Veritas Prep

On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $1, the second $100, and the third $1000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $1000?

(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2

Experts: Only Veritas Prep experts, please!
Since I'm not associated with Veritas, I've hidden the explanation below. It will be revealed if you use the quote function to respond.

[spoiler]There is only 1 way to win more than $1000: the contestant must assign all of the keys correctly. To win more than $1000, he must assign the $1000 key and at least 1 other key correctly. But it's impossible to assign exactly 2 keys correctly. Once 2 keys have been assigned correctly, the 3rd key must also be assigned correctly, because there will only 1 key and 1 box left, and this 1 remaining key and the 1 remaining box will have to match.

P($1000 box gets the correct key) = 1/3 (out of the 3 keys, only 1 is correct)
P($100 box gets the correct key) = 1/2 (out of the 2 remaining keys, only 1 is correct)
P($1 box gets the correct key) = 1/1 (1 key left, and it must be correct, since the incorrect keys have already been assigned to the other 2 boxes)

Since we need all of these events to happen, we multiply the fractions:

1/3 * 1/2 * 1/1 = 1/6.

The correct answer is C.[/spoiler]
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Wed Dec 01, 2010 4:46 am
GMATGuruNY wrote:
[spoiler]
P($1000 gets the correct key) = 1/3
P($100 box gets the correct key) = 1/2
P($1 box gets the correct key) = 1/1

[/spoiler]
Hi Guru,

This is the exact point where I was completely confused.

[spoiler]P($1000 gets the correct key) = 1/3
P($100 box gets the correct key) = 1/2
P($1 box gets the correct key) = 1/1

P(Win) = 1/6

What about the other case

P($1000 gets the correct key) = 1/3
P($1 box gets the correct key) = 1/2
P($100 box gets the correct key) = 1/1

P(Win) = 1/6

Thus total 1/3. [/spoiler]

Where am I going wrong? Why shall not we consider the latter?
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Master | Next Rank: 500 Posts
Posts: 437
Joined: Sat Nov 22, 2008 5:06 am
Location: India
Thanked: 50 times
Followed by:1 members
GMAT Score:580

by beat_gmat_09 » Wed Dec 01, 2010 5:05 am
My question is -
Why do we have to consider 1000$ in the beginning ?
Why not this - 100$ , 1000$,1$ or 1$,100$,1000$ or 1$,1000$, 100$
I think the contestant can have 1-3 successful shots.
Hope is the dream of a man awake
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Wed Dec 01, 2010 5:15 am
[spoiler]The order of the boxes is irrelevant. To win more than $1000, each of the keys must be assigned correctly. Let's consider the 3 keys that the contestant is placing:

P(first key is put in the correct box) = 1/3
P(second key is put in the correct box) = 1/2
P(third key is put in the correct box) = 1/1

1/3 * 1/2 * 1/1 = 1/6.

Does this help?[/spoiler]
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Wed Dec 01, 2010 6:03 am
GMATGuruNY wrote:[spoiler]The order of the boxes is irrelevant. To win more than $1000, each of the keys must be assigned correctly. Let's consider the 3 keys that the contestant is placing:

P(first key is put in the correct box) = 1/3
P(second key is put in the correct box) = 1/2
P(third key is put in the correct box) = 1/1

1/3 * 1/2 * 1/1 = 1/6.

Does this help?[/spoiler]
Is it because the there are 3 boxes and for getting more than a 1000$ all the boxes are bound to be selected correctly?
If the problem is Easy Respect it, if the problem is tough Attack it
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Wed Dec 01, 2010 9:41 am
shovan85 wrote:
Is it because the there are 3 boxes and for getting more than a 1000$ all the boxes are bound to be selected correctly?
[spoiler]Yes. Perhaps it would be easier to think of the problem this way:

3 boxes are placed before the contestant. Each box has a corresponding key.
To win more than 1000 dollars, the contestant must place the 3 keys in the correct order.
There are 3! = 6 ways to arrange the 3 keys. There is only 1 correct arrangement of the 3 keys.
Thus, P(correct arrangement) = 1/6.

Does this help?[/spoiler]
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Wed Dec 01, 2010 9:44 am
GMATGuruNY wrote:
shovan85 wrote:
Is it because the there are 3 boxes and for getting more than a 1000$ all the boxes are bound to be selected correctly?
[spoiler]Yes. Perhaps it would be easier to think of the problem this way:

3 boxes are placed before the contestant. Each box has a corresponding key.
To win more than 1000 dollars, the contestant must place the 3 keys in the correct order.
There are 3! = 6 ways to arrange the 3 keys. There is only 1 correct arrangement of the 3 keys.
Thus, P(correct arrangement) = 1/6.

Does this help?[/spoiler]
Great Explanation indeed :)
Thanks a lot
If the problem is Easy Respect it, if the problem is tough Attack it
Top
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Wed Dec 01, 2010 9:58 pm
limestone wrote:[spoiler]IMO: C[/spoiler]

This is somewhat similar to sibling problems.

The desired outcome is the contestant to win more than $1000.

So the key picked for the box that contains $1000 must match. However, the contestant needs to win more than $1000. So another key picked, either for the $100 box or $1 box, must match. And this also leads to the matching of the final box too (As the first two boxes have already had matched keys).

All three boxes must have their respectively matched keys.

My approach:
Pick the first box (whatever it is, never mind 'bout it), possibility for this box to have its matched key : 1/3 (three keys remain)
Pick the second box, possibility for 2nd box to have its matched key: 1/2 (two keys remain)
The final box, possibility for it to have a matched key is absolute, or 100%, or 1.

Total possibility: 1/3*1/2*1 = 1/6

Correct me if I'm wrong.
Brilliant Approach, 1/3 * 1/2 * 1 = 1/6
Great work,
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Wed Dec 01, 2010 10:01 pm
limestone wrote:[spoiler]IMO: C[/spoiler]

This is somewhat similar to sibling problems.

The desired outcome is the contestant to win more than $1000.

So the key picked for the box that contains $1000 must match. However, the contestant needs to win more than $1000. So another key picked, either for the $100 box or $1 box, must match. And this also leads to the matching of the final box too (As the first two boxes have already had matched keys).

All three boxes must have their respectively matched keys.

My approach:
Pick the first box (whatever it is, never mind 'bout it), possibility for this box to have its matched key : 1/3 (three keys remain)
Pick the second box, possibility for 2nd box to have its matched key: 1/2 (two keys remain)
The final box, possibility for it to have a matched key is absolute, or 100%, or 1.

Total possibility: 1/3*1/2*1 = 1/6

Correct me if I'm wrong.
Brilliant Approach, 1/3 * 1/2 * 1 = 1/6
Great work,
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Junior | Next Rank: 30 Posts
Posts: 24
Joined: Wed Apr 28, 2010 4:03 am

by junegmat221 » Sat Dec 04, 2010 5:29 am
Well, i didnt get much into the technical portions because we will be unsure as shovan did in the earlier posts.
Its always to better to approach from a Practical perspective.
We are given three keys namely A B C.
There are three boxes D E F( D is 1$ , E is 100$, F is 1000$)
Let us assume the right combination first
(AD, BE, CF are the only right combinations).

Here are the different scenarios.
ABC can be distributed in 3! ways. that means 6 essential possiblities.

D E F (Boxes)
============
A B C (This combination yields more than 1000$) = Favourable outcome
A C B (This combination yields just 1 $ since A Is properly inserted in D , whereas other two keys mismatch with boxes)
B A C (First two keys mismatch and third one yields 1000$ but its still just 1000$)
B C A (All Three mismatch and yields nothing)
C A B (All Three mismatch and yields nothing)
C B A (B alone matches with its box earning 100$)

So out of this all outcomes we get only the first output in our favourable outcomes.
So the probablility of getting more than 1000$ is 1/6
Top
Senior | Next Rank: 100 Posts
Posts: 86
Joined: Mon Sep 13, 2010 2:36 pm
Thanked: 29 times
Followed by:2 members
GMAT Score:710

by thebigkats » Sat Dec 18, 2010 1:42 pm
Hi:
I used simple probability to arrive the answer. Let me know if this approach is wrong and I just got lucky:

We absolutely need $1000 box to be open. Chances of that key getting into the right box is 1/3 (it could go into any box)
Assuming that this is actually in the right box, chances of other keys (2 left) in the right box (either of the two) is also 1/2 (only two possibilities - either both keys in right boxes or both in wrong boxes).

so to arrive >$1000 means that our possibility is 1/3 * 1/2 = 1/6

Certainly I could have gone down the path of listing out the combinations etc. but I got to this answer in few sec through probability
Top
Post new topic Post Reply